Vectors Proof Questions
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Either paper 5 marks
\(OAB\) is a triangle. \(P\) is the point on \(AB\) such that \(AP : PB = 3 : 1\).
\(\overrightarrow{OA} = 12\mathbf{a}\) and \(\overrightarrow{OB} = 12\mathbf{b}\).
(a)Find \(\overrightarrow{AB}\) in terms of \(\mathbf{a}\) and \(\mathbf{b}\).
(b)Find \(\overrightarrow{OP}\) in terms of \(\mathbf{a}\) and \(\mathbf{b}\).
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Either paper 3 marks
\(ABC\) is a triangle. The diagram is not drawn accurately.
\(D\) is a point such that \(\overrightarrow{AD} = -4\mathbf{a} + 6\mathbf{b}\).
Is \(BCD\) a straight line?
Show working to support your answer.
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Either paper 5 marks
\(OAB\) is a triangle.
\(\overrightarrow{OA} = 4\mathbf{a}\) and \(\overrightarrow{OB} = 2\mathbf{b}\).
\(X\) is the point on \(OA\) such that \(OX : XA = 1 : 1\), and \(Y\) is the point on \(OB\) such that \(OY : YB = 1 : 1\).
(a)Write \(\overrightarrow{XY}\) in terms of \(\mathbf{a}\) and \(\mathbf{b}\).
(b)Show that \(XY\) is parallel to \(AB\), and write down the ratio \(XY : AB\).
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Either paper 7 marks
\(OABC\) is a parallelogram. \(X\) is the midpoint of the diagonal \(AC\).
\(\overrightarrow{OA} = 8\mathbf{a}\) and \(\overrightarrow{OC} = 4\mathbf{b}\).
(a)Find \(\overrightarrow{AC}\) in terms of \(\mathbf{a}\) and \(\mathbf{b}\).
(b)Find \(\overrightarrow{OX}\) in terms of \(\mathbf{a}\) and \(\mathbf{b}\).
(c)Prove that \(O\), \(X\) and \(B\) lie on the same straight line.
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Either paper 1 mark
The points \(A\), \(B\) and \(C\) have position vectors, relative to an origin \(O\),
\(\overrightarrow{OA} = 2\mathbf{a} + 3\mathbf{b}\)
\(\overrightarrow{OB} = 5\mathbf{a} + 12\mathbf{b}\)
\(\overrightarrow{OC} = 9\mathbf{a} + 24\mathbf{b}\)
where \(\mathbf{a}\) and \(\mathbf{b}\) are not parallel.
Prove that \(A\), \(B\) and \(C\) lie on the same straight line.
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Either paper 6 marks
\(OABC\) is a trapezium. \(M\) is the midpoint of \(BC\).
\(\overrightarrow{OA} = 6\mathbf{a}\), \(\overrightarrow{AB} = 4\mathbf{b}\) and \(\overrightarrow{OC} = 12\mathbf{b}\).
(a)Write \(\overrightarrow{BC}\) in terms of \(\mathbf{a}\) and \(\mathbf{b}\).
(b)Write \(\overrightarrow{OM}\) in terms of \(\mathbf{a}\) and \(\mathbf{b}\).
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Either paper 7 marks
\(OABC\) is a parallelogram. \(P\) is the midpoint of \(AB\) and \(Q\) is the midpoint of \(BC\).
\(\overrightarrow{OA} = 8\mathbf{a}\) and \(\overrightarrow{OC} = 2\mathbf{b}\).
(a)Find \(\overrightarrow{AC}\) in terms of \(\mathbf{a}\) and \(\mathbf{b}\).
(b)Find \(\overrightarrow{PQ}\) in terms of \(\mathbf{a}\) and \(\mathbf{b}\).
(c)Show that \(PQ\) is parallel to \(AC\), and write down the ratio \(PQ : AC\).
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Either paper 4 marks
\(ABCD\) is a quadrilateral. \(E\), \(F\), \(G\) and \(H\) are the midpoints of \(AB\), \(BC\), \(CD\) and \(DA\).
\(\overrightarrow{AH} = \mathbf{a}\), \(\overrightarrow{AE} = 2\mathbf{b}\) and \(\overrightarrow{DG} = 5\mathbf{c}\).
Prove, using vectors, that \(EFGH\) is a parallelogram.
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Either paper 4 marks
\(ABC\) and \(ACD\) are triangles. \(k\) is a constant.
\(\overrightarrow{AB} = k\mathbf{a} + 6\mathbf{b}\)
\(\overrightarrow{AC} = 9\mathbf{a} + 8\mathbf{b}\)
\(\overrightarrow{AD} = 21\mathbf{a} + 12\mathbf{b}\)
(a)Show that \(\overrightarrow{CD} = 12\mathbf{a} + 4\mathbf{b}\)
(b)\(BCD\) is a straight line.
Calculate the value of \(k\). You must show your working.
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Either paper 4 marks
\(CDEF\) is a quadrilateral.
\(\overrightarrow{FE} = 2\mathbf{a}\), \(\overrightarrow{ED} = \mathbf{b}\) and \(\overrightarrow{DC} = -26\mathbf{a} + 3\mathbf{b}\).
\(P\) is the point such that \(CEP\) is a straight line and \(CE : EP = 1 : 3\).
Prove, using vectors, that \(CF\) is parallel to \(DP\).
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(a) \(12\mathbf{b} - 12\mathbf{a}\)
(b) \(3\mathbf{a} + 9\mathbf{b}\) - Yes: \(\overrightarrow{BC} = -2\mathbf{a} + 2\mathbf{b}\) and \(\overrightarrow{CD} = -4\mathbf{a} + 4\mathbf{b}\), so the coefficients are multiplied by \(2\) and \(2\), so CD is a multiple of BC. The two vectors are parallel and share the point C, so BCD is a straight line
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(a) \(\mathbf{b} - 2\mathbf{a}\)
(b) \(\overrightarrow{AB} = 2\mathbf{b} - 4\mathbf{a} = 2\left(\mathbf{b} - 2\mathbf{a}\right) = 2\,\overrightarrow{XY}\), so \(XY\) is parallel to \(AB\), and \(XY : AB = 1 : 2\) -
(a) \(4\mathbf{b} - 8\mathbf{a}\)
(b) \(4\mathbf{a} + 2\mathbf{b}\)
(c) \(\overrightarrow{OB} = 8\mathbf{a} + 4\mathbf{b} = 2\left(4\mathbf{a} + 2\mathbf{b}\right) = 2\,\overrightarrow{OX}\), so \(OB\) is parallel to \(OX\) and they share the point \(O\); therefore \(O\), \(X\) and \(B\) are on one straight line - \(\overrightarrow{AB} = 3\mathbf{a} + 9\mathbf{b}\) and \(\overrightarrow{BC} = 4\mathbf{a} + 12\mathbf{b} = \frac{4}{3}\,\overrightarrow{AB}\), so \(AB\) and \(BC\) are parallel and share the point \(B\); therefore \(A\), \(B\) and \(C\) are on one straight line
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(a) \(8\mathbf{b} - 6\mathbf{a}\)
(b) \(3\mathbf{a} + 8\mathbf{b}\) -
(a) \(2\mathbf{b} - 8\mathbf{a}\)
(b) \(\mathbf{b} - 4\mathbf{a}\)
(c) \(\overrightarrow{AC} = 2\mathbf{b} - 8\mathbf{a} = 2\left(\mathbf{b} - 4\mathbf{a}\right) = 2\,\overrightarrow{PQ}\), so \(PQ\) is parallel to \(AC\), and \(PQ : AC = 1 : 2\) - \(\overrightarrow{EH} = \mathbf{a} - 2\mathbf{b}\) and \(\overrightarrow{FG} = \mathbf{a} - 2\mathbf{b}\), so \(\overrightarrow{EH} = \overrightarrow{FG}\): \(EH\) and \(FG\) are equal and parallel, hence \(EFGH\) is a parallelogram
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(a) \(\overrightarrow{CD} = \overrightarrow{CA} + \overrightarrow{AD} = -\left(9\mathbf{a} + 8\mathbf{b}\right) + \left(21\mathbf{a} + 12\mathbf{b}\right) = 12\mathbf{a} + 4\mathbf{b}\)
(b) \(3\) - \(\overrightarrow{CF} = 24\mathbf{a} - 4\mathbf{b}\) and \(\overrightarrow{DP} = 78\mathbf{a} - 13\mathbf{b}\); these are \(-4\) and \(-13\) times \(-6\mathbf{a} +\mathbf{b}\), so \(CF\) is parallel to \(DP\)
Takes a couple of seconds the first time a sheet is made.